Question

An iron ball of mass $0.2kg$ is heated to${{100}^{0}}C$, when it is put in ice block at${{0}^{0}}C$, $25g$ ice is melt. The specific heat of iron in CGS unit is

& a)1 \\\ & b)0.1 \\\ & c)0.8 \\\ & d)0.08 \\\ \end{aligned}$$

Answer 1

Here Principle of Calorimetry is used which signifies that Heat lost by hot body is equal to heat gain by cold body. Here heat is lost by an iron ball when it is placed on ice and heat is gained by ice due to which it melts but temperature to ice remains at ${{0}^{0}}C$ because some of the ice is melted.

Complete step by step answer:
Let us assume that the specific heat of iron as c
Mass of iron ball (M) $=0.2Kg=200g$
Initial Temperature (T1) $={{100}^{0}}C$
When it is placed on ice block at ${{0}^{0}}C$ then only $25g$ ice melts i.e. rest of ice remain at ${{0}^{0}}C$ and final temperature of Iron ball becomes${{0}^{0}}C$.
Final Temperature (T2) =${{0}^{0}}C$
Mass of ice melts (m) $=25g$
We know that, Latent heat of fusion of ice = $80cal/g$
According to Principle of Calorimetry,
Heat Lost by Iron Ball = Heat gain by Ice
$Mc\Delta t=mL$
$Mc({{T}_{1}}-{{T}_{2}})=mL$
$200\times c\times (100-0)=25\times 80$
$200\times c\times 100=2000$
$c=\dfrac{2000}{20000}$
$c=0.1ca{{l}^{0}}C/g$
Correct answer is option b.

Addition Information: When heat is given and temperature of body changes then formula for heat absorbed used is $Q=Mc\Delta t$ where c is called specific heat of body
When heat is given and only the state of the body changes then the formula for heat absorbed is $Q=mL$ where L is latent heat of fusion/vaporization.

Note: Whenever heat is given to ice and it start melting its temperature remains at${{0}^{0}}C$till whole of the ice will melt and on further heating it its temperature increases .When heat is given and only state of body is changing i.e. temperature remains constant then only latent heat of Fusion/Vaporisation is used.